As we know that The Radiant flux density or Poynting vector
$$
\mathrm{S}=\frac{1}{\mu_0}(\mathrm{E} \times \mathrm{B})=\mathrm{c}^2 \varepsilon_0(\mathrm{E} \times \mathrm{B})\left[\because \mathrm{c}=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\right]
$$
Let electromagnetic waves be propagating along $\mathrm{x}$-axis. So its electric field vector of electromagnetic wave be along $\mathrm{y}$-axis and magnetic field vector be along z-axis. So that, and
$$
\begin{aligned}
\mathrm{E}_0 &=\mathrm{E}_0 \cos (\mathrm{kx}-\omega \mathrm{t}) \\
\mathrm{B} &=\mathrm{B}_0 \cos (\mathrm{kx}-\omega \mathrm{t}) \\
\mathrm{E} \times \mathrm{B} &=\left(\mathrm{E}_0 \cdot \mathrm{B}_0\right) \cos ^2(\mathrm{kx}-\omega \mathrm{t}) \\
\mathrm{S} &=\mathrm{c}^2 \varepsilon_0(\mathrm{E} \cdot \mathrm{B}) \\
&=\mathrm{c}^2 \varepsilon_0\left(\mathrm{E}_0 \cdot \mathrm{B}_0\right) \cos ^2(\mathrm{kx}-\omega \mathrm{t})
\end{aligned}
$$
So, average value of the magnitude of radiant flux density over complete cycle is
$$
\begin{aligned}
&\mathrm{S}_{\mathrm{av}}=\mathrm{c}^2 \varepsilon_0\left|\mathrm{E}_0 \cdot \mathrm{B}_0\right| \frac{1}{\mathrm{~T}} \int_0^{\mathrm{T}} \cos ^2(\mathrm{kx}-\omega \mathrm{t}) \mathrm{dt} \\
&=\mathrm{c}^2 \varepsilon_0 \mathrm{E}_0 \mathrm{~B}_0 \times \frac{1}{\mathrm{~T}} \times \frac{\mathrm{T}}{2}\left[\because \int_0^{\mathrm{T}} \cos ^2(\mathrm{kx}-\omega \mathrm{t}) \mathrm{dt}=\frac{\mathrm{T}}{2}, \mathrm{c}=\frac{\mathrm{E}_0}{\mathrm{~B}_0}\right]
\end{aligned}
$$
So, $S_{\mathrm{av}}=\frac{\mathrm{c}^2}{2} \varepsilon_0 \mathrm{E}_0\left(\frac{\mathrm{E}_0}{\mathrm{c}}\right)$
$$
S_{\mathrm{av}}=\frac{\mathrm{c}}{2} \varepsilon_0 \mathrm{E}_0^2=\frac{\mathrm{c}}{2} \times \frac{1}{\mathrm{c}^2 \mu_0} \mathrm{E}_0^2\left[\because \mathrm{c}=\frac{1}{\sqrt{\mu_0 \varepsilon_0}} \text { or } \varepsilon_0=\frac{1}{\mathrm{c}^2 \mu_0}\right]
$$
So, $\mathrm{S}_{\mathrm{av}}=\frac{\mathrm{E}_0^2}{2 \mu_0 \mathrm{c}}$
Hence Proved.